Question

Supposse $a,b,c$ are in A.P. and $a^2,b^2,c^2$ are in G.P. If $a<b<c$ and $a+b+c=\frac{3}{2}$, then the value of $a$ is

• A. $\frac{1}{2\sqrt{2}}$
• B. $\frac{1}{2\sqrt{3}}$
• C. $\frac{1}{2}-\frac{1}{\sqrt{3}}$
• D. $\frac{1}{2}-\frac{1}{\sqrt{2}}$

Answer 1

The correct option is D $\frac{1}{2}-\frac{1}{\sqrt{2}}$

Given that $a,b,c$ are in A.P.
$\Rightarrow 2b=a+c$
$\text{But given}a+b+c=\frac{3}{2}\Rightarrow 3b=\frac{3}{2}\Rightarrow b=\frac{1}{2}andthena+c=1$
Again $a^2,b^2,c^2,areinG.P.\Rightarrow b^4=a^2{c}^2\Rightarrow b^2=\pm ac\Rightarrow ac=\frac{1}{4}or-\frac{1}{4}$
and $a+c=1...\left(i\right)$
Considering $a+c=1andac=\frac{1}{4}\Rightarrow (a-c{)}^2=1-1=0\Rightarrow a=c$, but
$a\ne c$ as given that $a<b<c<$
$\therefore \text{We consider}a+c=1andac=-\frac{1}{4}\Rightarrow (a-c{)}^2=1+1=2\Rightarrow a-c=\pm \sqrt{2}$
but $a<c\Rightarrow a-c=-\sqrt{2}...\left(ii\right)$
Solving (i) and (ii) we get $a=\frac{1}{2}-\frac{1}{\sqrt{2}}$